Table Of ContentORTHOGONALITY OF HOMOGENEOUS GEODESICS ON THE
TANGENT BUNDLE
1
1 R.CHAVOSHKHATAMY∗
0
2
n
a Abstract. Letξ=(G×KG/K,ρξ,G/K,G/K)betheassociatedbundleand
J τ = (T ,π ,G/K,Rm) be the tangent bundle of special examples
G/K G/K G/K
0 ofodddimensionsolvableLiegroupsequippedwithleftinvariantRiemannian
1 metric. Inthispaperweprovesomeconditionsabouttheexistenceofhomoge-
] neousgeodesiconthebasespaceofτG/K andhomogeneous(geodesic)vectors
G onthefiberspaceofξ .
D
.
h
t
a 1. Introduction and preliminaries
m
[ LetGbeaconnectedLiegroupandK beaclosedsubgroupofG. Thesetofleft
1 cosetsofK inGisdenotedbyG/K andcanbe givenaunique differentiablestruc-
v
ture ([6], vol.II, chap.2), and hence M = G/K is called a homogeneous manifold.
8
2 When a Lie group G acts transitively isometric on a Riemannian manifold M, we
9
can identify M with the set G/K of left cosets of the isotropy group K of a point
1
. x0 ∈M. The point x0 is called the origin of M. Let ▽ be an affine connection on
1
0 M = G/K and let ▽ be invariant under the natural action of T : G×M −→ M.
1
Then a geodesic γ : I −→ M is called a homogeneous geodesic if, there exists a
1
: 1-parameter subgroup t−→exptX , t∈R, of G with X ∈G=TeG such that
v
i
X γ(t)=T(exptX,x0).
r
a
Where γ(0)=x0 ∈M, and exp:G →G is the exponential map [9].
Definition 1.1. Avector06=X ∈G iscalledahomogeneous vector (orgeodesic
vector), if the curve γ(t)=(exptX)(x0) is a geodesic on M =G/K [9].
The following result can be found in [7], proposition 1.
2000 Mathematics Subject Classification. 53C30,53C22.
Keywordsandphrases. Hadamardmatrices,tangentbundle,homogeneousmanifold,solvable
Liegroup,homogeneous (geodesic) vector.
∗Department of Mathematics, Islamic Azad UniversityTabriz Branch,
Tabriz-IRAN
E-mail: [email protected], r [email protected] .
1
2 R.CHAVOSHKHATAMY
Any homogeneous Riemannian manifold G/K has the reductive decomposition
of the form
G =M+K
where M⊂G is a vector subspace, such that Ad(K)(M)⊂M.
LetM =G/K be a RiemannianmanifoldandG =M+K,its reductivedecom-
position. Thenthenaturalmapφ:G−→G/K =M inducesalinearepimorphism
(dφ) : T G −→ T M, and the vector space M can be identified with T M. If
e e x0 x0
C is a scalar product on M induced by the scalar product on T M, then the
x0
following lemma holds (see [9], proposition 2.1).
Lemma1.2. IfX belongstoG,let[X,Y]M andXM bethecomponentsof[X,Y]
and X in M withrespect toreductive decomposition, then X is homogeneous vector
(or geodesic vector) iff
C(XM,[X,Y]M)=0 ∀Y ∈G.
Proposition1.3.([10]). Afinitefamily{γ1,γ2,...,γn}ofhomogeneousgeodesics
throughx ∈M isorthogonal(respectively,linearlyindependent)iftheM-component
o
of the corresponding homogeneous vectors are orthogonal (respectively, linearly in-
dependent).
Let ℘ = (P,π,B,G) be a smooth fiber bundle. A pair (℘,T) is called a
(smooth)principal bundle with structure group G, if T : P ×G 7−→ P is a right
action of G on P and ℘ admits a coordinate representation {(U ,ψ )} such that
α α
ψ (x,ab)=ψ (x,a)b, x∈U , a,b∈G,
α α α
(see [6], vol.II, chap.V).
Let ℘ = (P,π,B,G) be a principal bundle and F be a differentiable manifold.
Consider the left action Q, of G on the product manifold P ×F given by
Q (z,y)=(z,y)a=(za,a−1y) z ∈P,y ∈F,a∈G.
a
The set of orbits of this action is denoted by P × F and
G
q :P ×F →P × F
G
ORTHOGONALITY OF HOMOGENEOUS GEODESICS ... 3
will denote the corresponding projection, i.e., q(z,y) is the orbit through (z,y).
The map q determines a map ρ :P × F →B such that,
ξ G
ρ ◦q =π◦π .
ξ p
Where, π : P ×F → P is the canonical projection and π : P → B is the bundle
p
map.
There is a unique smoothstructure onP × F, suchthat ξ =(P × F,ρ ,B,F)is
G G ξ
a smooth fiber bundle (see [6], vol.II, chap.V, sec.2).
Definition 1.4. Thefiberbundleξ =(P× F,ρ ,B,F),iscalledtheassociated
G ξ
bundle with ℘=(P,π,B,G).
Let K be a closed subgroup of G. The principal fiber bundle ℑ=(G,π,G/K,K),
is called homogeneous bundle, (See [3]).
Let ℑ = (G,π,G/K,K) be a fiber bundle with group structure K, and let G
be a connected Lie group and K a closed subgroup of G, (see [1], definition 2.2).
We take the Lie algebras G and K of G and K respectively, in [1] and [2], we
proved some relations between the homogeneous vector in the fiber space of the
associated bundle, ξ = (G× G/K,ρ ,G/K,G/K) and the homogeneous geodesic
K ξ
in the base spaceof a principalhomogeneousbundle ℑ=(G,π,G/K,K).In[3], we
considerthehomogeneousbundleℑ=(G,π,G/K,K)andthetangentbundleτ
G/K
of M = G/K, and give some results about the existence of homogeneous vectors
on the fiber space of τ , for both cases of G semisimple and weakly semisimple
G/K
Lie group.
Now, we investigate the existence of mutually orthogonal linearly independent
homogeneousgeodesicsinthe base spaceofthe tangentbundle τ ofhomogeneous
G
Riemannian manifold G given in theorem 2.2.
2. Main results
Letℑ=(G,π,G/K,K)beaprincipalhomogeneousbundle,withtheassociated
bundle ξ =(G× K,ρ ,G/K,G/K). Let
K ξ
G be the matrix group of all matrices of the form
ez0 0 ... 0 x0
0 ez1 ... 0 x1
... ... ... ... ...
0 0 ... ezn x
n
0 0 ... 0 1
4 R.CHAVOSHKHATAMY
where, (x0,x1,··· ,xn,z1··· ,zn) ∈ R2n+1. The Lie group G is unimodular and
solvable (see [8], pp.134-136),with the left invariant Riemannian metric
n n
g = e−2zidx2+λ2 dz dz .
X i X k j
i=0 k,j=0
Where λ 6= 0 is a constant. Then G is a homogeneous Riemannian manifold
with the origin at (0,0,··· ,0) ([8], p.134).
Let G = M+K be the reductive decomposition of G , then K= 0, and hence
G =M.
In [3], we prove the following lemma
Lemma 2.1. Let ℑ=(G,π,G/K,K), be a homogeneous bundle. Then
ξ =(G× G/K,ρ ,G/K,G/K),
K ξ
is the associated bundle of ℑ=(G,π,G/K,K).
By lemma 2.1, we can take ξ =(G×M,ρ ,G,M), be the associated bundle of
ξ
ℑ=(G,π,G/K,K).
In [4], we let G be a 3-dimensional solvable Lie group, given in [8], pp.134, and
prove some results about the existence of homogeneous vectors on the fiber space
of τ and ξ.
G/K
In [5], we extend theorems 5.6 and 5.7 in [4], and give the following theorem, for
the odd dimensional solvable Lie group.
Theorem2.2.([5]). Letℑ=(G,π,G/K,K), beaprincipalhomogeneous bundle
and ξ =(G× G/K,ρ ,G/K,G/K),betheassociatedbundleofℑ=(G,π,G/K,K).
K ξ
If G is the matrix group of all matrices of the form
ez0 0 ... 0 x0
0 ez1 ... 0 x1
... ... ... ... ...
0 0 ... ezn x
n
0 0 ... 0 1
where (x0,x1,··· ,xn,z1··· ,zn)∈R2n+1 and z0 =−(z1+z2+···+zn). Then
a vector V in the fiber space of ξ is a homogenous (geodesic ) if and only if its
components
(x0,x1,··· ,xn,z1,··· ,zn)
satisfy the following conditions
ORTHOGONALITY OF HOMOGENEOUS GEODESICS ... 5
x0(z1+z2+···+zn)=0 x1z1 =0,··· ,xnz1 =0
x2−x2 =0,··· ,x2−x2 =0.
0 1 0 n
In the proof of the Theorem 5.3 in [3], we give a strong isomorphism between
the tangent bundle
τ =(T ,π ,G/K,Rm)
G/K G/K G/K
and the associated bundle
ξ =(G× G/K,ρ ,G/K,G/K),
K ξ
then under hypothesis of theorem 2.2 there is a strong isomorphism between,the
associated bundle
ξ =(G×M,ρ ,G,M)
ξ
and the tangent bundle
τ =(T ,π ,G,R2n+1)
G G G
so we have,
Corollary 2.3.([5]). With hypothesis of theorem 2.2, let
τ =(T ,π ,G,R2n+1)
G G G
be the tangent bundle of the homogeneous Riemannian manifold G. Then a vec-
tor W in R2n+1 is a homogeneous vector (under isomorphism), if and only if its
component (x0,x1,··· ,xn,z1··· ,zn) satisfy the following conditions
x0(z1+z2+···+zn)=0 x1z1 =0,··· ,xnz1 =0
x2−x2 =0,··· ,x2−x2 =0.
0 1 0 n
In[3],theorem5.4. wegiveasubspaceofG′suchthatallmemberofthissubspace
are homogeneous vectors, and by strong isomorphism between τ and ξ we can
G/K
findasubspaceofRm (underisomorphism)suchthatallmembersofthissubspace
are homogeneous vectors,
In the following theorem, we consider the tangent bundle
τ =(T ,π ,G,R2n+1)
G G G
of the homogeneous Riemannian manifold G in theorem 2.2, and give structure
of all subspaces of R2n+1 such that all their members are homogeneous vectors.
Theorem 2.4.([5]). Let
τ =(T ,π ,G,R2n+1)
G G G
6 R.CHAVOSHKHATAMY
be the tangent bundle of the homogeneous Riemannian manifold G, (given in theo-
rem2.2). Thenallhomogeneous vectorsaredecomposedintoann-dimensionvector
subspace W in R2n+1 and 2n, one-dimension vector subspace in R2n+1 generated
by all vectors of the form X0±X1±···±Xn.
By proposition1.3 andTheorem2.4we have,the followingresultaboutlinearly
independence of homogeneous geodesics on the base space of τ
G
Corollary 2.5.([5]). With hypothesis of theorem 2.2, the tangent bundle
τ =(T ,π ,G,R2n+1)
G G G
admits 2n+1 linearly independent homogeneous geodesics through the origin {e}
of the base space of τ .
G
Now, we investigate orthogonality of homogeneous vectors on the fiber space of
tangent bundle,
τ =(T ,π ,G,R2n+1).
G G G
In [3] we provesome conditions aboutexistence andorthogonalityof homogeneous
vectors for both cases of G semisimple and weakly semisimple. For example in
theorem 5.3 in [3], we prove that if G is a semisimple Lie group then there are m
orthogonal homogeneous vectors on the fiber space of the tangent bundle,
τ =(T ,π ,G/K,Rm)
G/K G/K G/K
Inthefollow,wewanttogetsomeconditionsaboutlinearlyindependentandor-
thogonality of homogeneous vectors on the fiber space and homogeneous geodesics
on the base space of the tangent bundle of the homogeneous Riemannian manifold
G (given in theorem 2.2). For this we need to considering to relations between
orthogonality of homogeneous vectors and the Hadamard matrices.
Definition 2.6. A Hadamard matrix of order k is k×k square matrix whose
entries are all equal to ±1, and such that A.At =kI , where I is the unit matrix.
k k
The condition A.At =kI , in definition 2.6, implies that the k rows or columns
k
of a Hadamard matrix represent orthogonal k-tuples, with all entries equal to +1
or -1, we can use this fact for considering to structure of Hadamard matrices and
orthogonality of homogeneous (geodesic) vectors.
Lemma 2.7. Let τ = (T ,π ,G,R2n+1) be the tangent bundle of the homo-
G G G
geneous Riemannian Lie group G of all matrices of the form
ORTHOGONALITY OF HOMOGENEOUS GEODESICS ... 7
ez0 0 ... 0 x0
0 ez1 ... 0 x1
... ... ... ... ...
0 0 ... ezn x
n
0 0 ... 0 1
where (x0,x1,··· ,xn,z1··· ,zn)∈R2n+1 and z0 =−(z1+z2+···+zn) ,
then;
(i) If (n+1) is odd, then there are not any two mutually orthogonal (n+1)-tuples
with all entries equal to ±1.
(ii)If (n+1) is even and not divisible by 4, then there are exactly two mutually
orthogonal (n+1)-tuples with all entries equal to ±1.
Proof. Let τ =(T ,π ,G,R2n+1) be the tangentbundle of the homogeneous
G G G
Riemannian manifoldG (givenin theorem2.2). By Corollary2.3, andtheorem2.4
a vector w in R2n+1 is a homogeneous (geodesics) vector (under isomorphism), if
and only if
A) w ∈W =span(Z1,Z2,··· ,Zn)
i=n
B) w= x X and x2−x2 =0,··· ,x2−x2 =0.
X i i 0 1 0 n
i=0
As concernshomogeneous(geodesics)vectorsoftype (B),they areallgenerated
by the vectors of the form X0+ǫ1X1+···+ǫnXn, where ǫi ∈{1,−1}. Therefore,
the problem of finding mutually orthogonal geodesics vectors of type (B) is equiv-
alent to the algebraic problem of finding (n+1)-tuples, with all entries equal to
±1, which are mutually orthogonal with respect to the standard scaler product in
Rn+1.
Let (n+1) be odd number and W1 and W2 be two (n+1)-tuples with all entries
equal to ±1. The scaler product of W1 and W2 is the sum of the products of their
entries and all such products are equal to ±1. By hypotheses, (n+1) is odd, then
sum of the products of their entries dose not vanish, so W1 and W2 can not be
orthogonal, so we obtain (i). For the second statement of the lemma, we spouse
that (n +1) = 2m, where m is odd, let V1 and V2 be two (n +1)-tuples with
all entries equal to ±1, such that V1 = (1,1,··· ,1) and V2 = (−1,1,···−1,1),
then V1 and V2 are orthogonal. Now, we spouse that V, W, Z, are three mutu-
ally orthogonal (n+1)-tuples with all entries equal to ±1. Then, we compute the
scaler product of V, W and Z by V. In this way, we can obtain three mutually
8 R.CHAVOSHKHATAMY
orthogonal (n + 1)-tuples vectors V′, W′, Z′ such that all entries equal to ±1.
If we take V′ = (−1,−1,··· ,−1), then by orthogonality of V′ and W′, W′ has
exactly m entries equal to −1 and exactly m entries equal to 1. We then multi-
ply, component by component, and applying a fixed permutation of the all entries
for mutually orthogonal (n+1)-tuples vectors V′, W′, Z′, such that this applica-
tions will preserve the orthogonality of V′, W′, Z′. By this way, we can obtain
W′ =(1,1,··· ,1,−1,−1,··· ,−1), but m is odd and the orthogonality of V′, W′,
Z′ is imposable,thisgivesacontradiction,andthe proofofthe lemma iscomplete.
(cid:3)
Before starting some additional results, we recall the fact that A.At = kI , in
k
definition 2.6 implies that the k rows or columns of a Hadamard matrix represent
orthogonal k-tuples, with all entries equal to +1 or -1, for the case n+1 be divis-
ible by 4, the problem related to algebraic problem of the existence of Hadamard
matrices of order n+1. Therefore, we get at once the following proposition.
Proposition 2.8.With hypothesis of lemma 2.7, let n+1 be divisible by 4, then
R2n+1 admits n+1 mutuallyorthogonal (n+1)-tuplesvectors with all entries equal
to ±1, if and only if, there exists a Hadamard matrices of order n+1.
Now,wecanprovethefollowingtheoremaboutthelinearlyindependentandthe
maximum number of the orthogonal homogeneous (geodesic) vectors on the fiber
space of
τ =(T ,π ,G,R2n+1).
G G G
Theorem 2.9. Let τ =(T ,π ,G,R2n+1) be the tangent bundle of the homo-
G G G
geneous Riemannian Lie group G, (given in theorem 2.2 and lemma 2.7) then;
(i)There are 2n+1 linearly independent homogeneous (geodesics) vectors in the
fiber space of through the τ .
G
(ii) The maximum number of the orthogonal homogeneous (geodesic) vectors on the
fiber space of τ is n+1, in the case that n+1 is odd.
G
(ii)The maximum number of the orthogonal homogeneous (geodesic) vectors on the
fiber space of τ , is n+2, in the case that n+1 is even and not divisible by 4.
G
(iv)The maximum number of the orthogonal homogeneous (geodesic) vectors on the
fiber space of τ , is 2n+1, in the case that n+1 is even and divisible by 4 and
G
ORTHOGONALITY OF HOMOGENEOUS GEODESICS ... 9
there exists a Hadamard matrices of order n+1.
Proof. Theorem 2.4 and corollary 2.3, conclude the fist part of theorem, it is
easy to see that, there exist n+1 linearly independent homogeneous (geodesics)
vectors of type (B), ( see proof of lemma 2.7), then there are 2n+1 linearly inde-
pendent homogeneous (geodesics) vectors in the fiber space of τ .
G
The second and the third part of the theorem follows from (i) and (ii), in lemma
2.7. Finally, as an immediate consequence from proposition 2.8, we obtain (iv).(cid:3)
Byproposition1.3andtheorem2.9we completecorollary2.5aboutthe number
of linearly independent homogeneous geodesics through origin of the base space of
τ .
G
Corollary 2.10. Let τ = (T ,π ,G,R2n+1) be the tangent bundle of the ho-
G G G
mogeneous Riemannian Lie group G, (given in theorem 2.2 and lemma 2.7) then;
(i)There are 2n+1 linearly independent homogeneous geodesics vectors through
the origin {e} of the base space of τ .
G
(ii) The maximum number of the orthogonal homogeneous geodesic through the ori-
gin {e} of the base space of τ , is n+1, in the case that n+1 is odd.
G
(ii)The maximum number of the orthogonal homogeneous geodesic through the ori-
gin {e} of the base space of τ , is n+2, in the case that n+1 is even and not
G
divisible by 4.
(iv)The maximum number of the orthogonal homogeneous geodesic through the ori-
gin {e} of the base space of τ , is 2n+1, in the case that n+1 is even and divisible
G
by 4 and there exists a Hadamard matrices of order n+1.
Acknowledgement
The author would like to express his appreciation of professor O. Kowalski for
his invaluable suggestions and also professor M. Toomanian for his constructive
comments.
The author was supported by the funds of the Islamic Azad University- Tabriz
Branch, (IAUT).
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